Proportion and similarity question

[Jasonbot]

Hi everyone. COuld you guys please help me with this simple geometry problem?

In triangleADC, angleD=90*, DB perpendicular AC, BK||AD and 2DK=KC
Using triangleDAB ||| triangleCAD (already proven using AAA) prove that 2DA=sqrt3.BC

thanks,
Jason

 

[Jasonbot]

Don't they teach geometry in foreign countries?

 

[swiftaw]

OK,

AB^2 + BD^2 = AD^2
BC^2 + BD^2 = CD^2

Putting these together we get that

AB^2 + CD^2 - BC^2 = AD^2 (*)

Now, by similarity we have that AC = 3AB, so AC^2 = 9AB^2.

Using that fact (*) becomes
(1/9)AC^2 + CD^2 - BC^2 = AD^2

But AC^2 = AD^2 + CD^2, thus

(1/9)(AD^2 + CD^2) + CD^2 - BC^2 = AD^2

Re-arranging we get
(10/9)CD^2 - BC^2 = (8/9)AD^2

Now, use the fact that CD^2 = AC^2 - AD^2 we get

(10/9)(AC^2 - AD^2) - BC^2 = (8/9)AD^2
or, in other words
2AD^2 = (10/9)AC^2 - BC^2 (**)

But, AC = (3/2)BC, so substituting into (**) gets us

2AD^2 = (10/9)(9/4)BC^2 - BC^2 = (6/4)BC^2

Doubling both sides

4AD^2 = (12/4)BC^2 = 3BC^2

Taking square roots

2AD = sqrt(3)*BC

 

[Jasonbot]

Thank you so much! I have no idea how you see these things but I'm sure Ill get it 😕 DO you look at what you're supposed to prove and work backwards?

 

[swiftaw]

Thank you so much! I have no idea how you see these things but I'm sure Ill get it 😕 DO you look at what you're supposed to prove and work backwards?

No, that is bad form. The only thing I looked in the final answer was that it only contained AD and BC. I started by writing down things I knew to be true and then tried to combine them in such a way to leave only AD and BC.